5Solving Systems of Linear Equations - Big Ideas …...5Solving Systems of Linear Equations 5.1 Solving Systems of Linear Equations by Graphing 5.2 Solving Systems of Linear Equations

5 Solving Systems of Linear Equations5.1 Solving Systems of Linear Equations by Graphing5.2 Solving Systems of Linear Equations by Substitution5.3 Solving Systems of Linear Equations by Elimination5.4 Solving Special Systems of Linear Equations5.5 Solving Equations by Graphing5.6 Linear Inequalities in Two Variables5.7 Systems of Linear Inequalities

Fishing (p. 265)

Pets (p. 250)

Drama Club (p. 228)

Fruit Salad (p. 255)

Roofing Contractor (p. 222)

Mathematical Thinking: Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace.

Drama Club (p 228)

RRo ffofiing CCo tntra tctor (((p. 2222 )2)2)

y q

FiFi hshiing ((p. 26265)5)

SEE the Big Idea

FFr iuitt SSalladd ((p 25255)5)

217

Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyGraphing Linear Functions (A.3.C)

Example 1 Graph 3 + y = 1 — 2 x.

Step 1 Rewrite the equation in slope-intercept form.

y = 1 — 2 x − 3

Step 2 Find the slope and the y-intercept.

m = 1 — 2 and b = −3

Step 3 The y-intercept is −3. So, plot (0, −3).

Step 4 Use the slope to find another point on the line.

slope = rise — run

= 1 — 2

Plot the point that is 2 units right and 1 unit up from (0, −3). Draw a line through the two points.

Graph the equation.

1. y + 4 = x 2. 6x − y = −1 3. 4x + 5y = 20 4. −2y + 12 = −3x

Solving and Graphing Linear Inequalities (7.10.B, A.5.B)

Example 2 Solve 2x − 17 ≤ 8x − 5. Graph the solution.

2x − 17 ≤ 8x − 5 Write the inequality.

+ 5 + 5 Add 5 to each side.

2x − 12 ≤ 8x Simplify.

− 2x − 2x Subtract 2x from each side.

−12 ≤ 6x Simplify.

−12 — 6 ≤ 6x —

6 Divide each side by 6.

−2 ≤ x Simplify.

The solution is x ≥ −2.

0−5 −4 −3 −2 −1 1 32

x ≥ –2

Solve the inequality. Graph the solution.

5. m + 4 > 9 6. 24 ≤ −6t 7. 2a − 5 ≤ 13

8. −5z + 1 < −14 9. 4k − 16 < k + 2 10. 7w + 12 ≥ 2w − 3

11. ABSTRACT REASONING The graphs of the linear functions g and h have different slopes. The value of both functions at x = a is b. When g and h are graphed in the same coordinate plane, what happens at the point (a, b)?

x

y2

−1

−4

42−2−4

1(0, −3)2

218 Chapter 5 Solving Systems of Linear Equations

Using a Graphing Calculator

Mathematical Mathematical ThinkingThinking

Monitoring ProgressMonitoring ProgressUse a graphing calculator to fi nd the point of intersection of the graphs of the two linear equations.

1. y = −2x − 3 2. y = −x + 1 3. 3x − 2y = 2y = 1 — 2 x − 3 y = x − 2 2x − y = 2

Mathematically profi cient students select tools, including real objects, manipulatives, paper and pencil, and technology as appropriate, and techniques, including mental math, estimation, and number sense as appropriate, to solve problems. (A.1.C)

Core Core ConceptConceptFinding the Point of Intersection You can use a graphing calculator to fi nd the point of intersection, if it exists, of the graphs of two linear equations.

1. Enter the equations into a graphing calculator.

2. Graph the equations in an appropriate viewing window, so that the point of intersection is visible.

3. Use the intersect feature of the graphing calculator to fi nd the point of intersection.


Use a graphing calculator to fi nd the point of intersection, if it exists, of the graphs of the two linear equations.

y = − 1 — 2 x + 2 Equation 1

y = 3x − 5 Equation 2

SOLUTION

The slopes of the lines are not the same, so you know that the lines intersect. Enter the equations into a graphing calculator. Then graph the equations in an appropriate viewing window.

Use the intersect feature to fi nd the point of intersection of the lines.

The point of intersection is (2, 1).

−6

−4

4

66

y = − x + 212

y = 3x − 5

−6

−4

4

6

IntersectionX=2 Y=1

Section 5.1 Solving Systems of Linear Equations by Graphing 219

5.1 Solving Systems of Linear Equations by Graphing

Writing a System of Linear Equations

Work with a partner. Your family opens a bed-and-breakfast. They spend $600 preparing a bedroom to rent. The cost to your family for food and utilities is $15 per night. They charge $75 per night to rent the bedroom.

a. Write an equation that represents the costs.

Cost, C (in dollars)

= $15 per night ⋅

Number of nights, x + $600

b. Write an equation that represents the revenue (income).

Revenue, R (in dollars)

= $75 per night ⋅

Number of nights, x

c. A set of two (or more) linear equations is called a system of linear equations. Write the system of linear equations for this problem.

Essential QuestionEssential Question How can you solve a system of linear equations?

Using a Table or Graph to Solve a System

Work with a partner. Use the cost and revenue equations from Exploration 1 to determine how many nights your family needs to rent the bedroom before recovering the cost of preparing the bedroom. This is the break-even point.

a. Copy and complete the table.

b. How many nights does your family need to rent the bedroom before breaking even?

c. In the same coordinate plane, graph the cost equation and the revenue equation from Exploration 1.

d. Find the point of intersection of the two graphs. What does this point represent? How does this compare to the break-even point in part (b)? Explain.

Communicate Your AnswerCommunicate Your Answer 3. How can you solve a system of linear equations? How can you check your

solution?

4. Solve each system by using a table or sketching a graph. Explain why you chose each method. Use a graphing calculator to check each solution.

a. y = −4.3x − 1.3 b. y = x c. y = −x − 1 y = 1.7x + 4.7 y = −3x + 8 y = 3x + 5

x (nights) 0 1 2 3 4 5 6 7 8 9 10 11

C (dollars)

R (dollars)

APPLYING MATHEMATICS

To be profi cient in math, you need to identify important quantities in real-life problems and map their relationships using tools such as diagrams, tables, and graphs.

A.2.IA.3.FA.3.GA.5.C

TEXAS ESSENTIAL KNOWLEDGE AND SKILLS


5.1 Lesson What You Will LearnWhat You Will Learn Check solutions of systems of linear equations. Solve systems of linear equations by graphing.

Use systems of linear equations to solve real-life problems.

Systems of Linear Equationssystem of linear equations, p. 220solution of a system of linear equations, p. 220

Previouslinear equationordered pair

Core VocabularyCore Vocabullarry

Checking Solutions

Tell whether the ordered pair is a solution of the system of linear equations.

a. (2, 5); x + y = 7 Equation 12x − 3y = −11 Equation 2

b. (−2, 0); y = −2x − 4 Equation 1y = x + 4 Equation 2

SOLUTION

a. Substitute 2 for x and 5 for y in each equation.

Equation 1 Equation 2

x + y = 7 2x − 3y = −11

2 + 5 =?

7 2(2) − 3(5) =?

−11

7 = 7 ✓ −11 = −11 ✓ Because the ordered pair (2, 5) is a solution of each equation, it is a solution of

the linear system.

b. Substitute −2 for x and 0 for y in each equation.


y = −2x − 4 y = x + 4

0 =?

−2(−2) − 4 0 =?

−2 + 4

0 = 0 ✓ 0 ≠ 2 ✗ The ordered pair (−2, 0) is a solution of the fi rst equation, but it is not a solution

of the second equation. So, (−2, 0) is not a solution of the linear system.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comTell whether the ordered pair is a solution of the system of linear equations.

1. (1, −2); 2x + y = 0−x + 2y = 5

2. (1, 4); y = 3x + 1y = −x + 5

READINGA system of linear equations is also called a linear system.

A system of linear equations is a set of two or more linear equations in the same variables. An example is shown below.

x + y = 7 Equation 1

2x − 3y = −11 Equation 2

A solution of a system of linear equations in two variables is an ordered pair that is a solution of each equation in the system.


Solving Systems of Linear Equations by GraphingThe solution of a system of linear equations is the point of intersection of the graphs of the equations.

Core Core ConceptConceptSolving a System of Linear Equations by GraphingStep 1 Graph each equation in the same coordinate plane.

Step 2 Estimate the point of intersection.

Step 3 Check the point from Step 2 by substituting for x and y in each equation of the original system.

Solving a System of Linear Equations by Graphing

Solve the system of linear equations by graphing.

y = −2x + 5 Equation 1

y = 4x − 1 Equation 2

SOLUTION

Step 1 Graph each equation.

Step 2 Estimate the point of intersection. The graphs appear to intersect at (1, 3).

Step 3 Check your point from Step 2.


y = −2x + 5 y = 4x − 1

3 =?

−2(1) + 5 3 =?

4(1) − 1

3 = 3 ✓ 3 = 3 ✓ The solution is (1, 3).

Check

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comSolve the system of linear equations by graphing.

3. y = x − 2 4. y = 1 — 2 x + 3 5. 2x + y = 5

y = −x + 4 y = − 3 — 2 x − 5 3x − 2y = 4

REMEMBERNote that the linear equations are in slope-intercept form. You can use the method presented in Section 3.5 to graph the equations.

x

y

2

42−2−4

(1, 3)y = −2x + 5

1 3)

y = 4x − 1

−1

−6

−2

6

6

y = 4x − 1y = −2x + 5

IntersectionX=1 Y=3


Modeling with Mathematics

A roofi ng contractor buys 30 bundles of shingles and 4 rolls of roofi ng paper for $1040. In a second purchase (at the same prices), the contractor buys 8 bundles of shingles for $256. Find the price per bundle of shingles and the price per roll of roofi ng paper.

SOLUTION

1. Understand the Problem You know the total price of each purchase and how many of each item were purchased. You are asked to fi nd the price of each item.

2. Make a Plan Use a verbal model to write a system of linear equations that represents the problem. Then solve the system of linear equations.

3. Solve the Problem

Words 30 ⋅ Price per bundle + 4 ⋅ Price per roll

= 1040

8 ⋅ Price per bundle + 0 ⋅ Price per roll = 256 Variables Let x be the price (in dollars) per bundle and let y be the

price (in dollars) per roll.

System 30x + 4y = 1040 Equation 1

8x = 256 Equation 2

Step 1 Graph each equation. Note that only the fi rst quadrant is shown because x and y must be positive.

Step 2 Estimate the point of intersection. The graphs appear to intersect at (32, 20).

Step 3 Check your point from Step 2.


30x + 4y = 1040 8x = 256

30(32) + 4(20) =?

1040 8(32) =?

256

1040 = 1040 ✓ 256 = 256 ✓ The solution is (32, 20). So, the price per bundle of shingles is $32, and the

price per roll of roofi ng paper is $20.

4. Look Back You can use estimation to check that your solution is reasonable. A bundle of shingles costs about $30. So, 30 bundles of shingles and 4 rolls of roofi ng paper (at $20 per roll) cost about 30(30) + 4(20) = $980, and 8 bundles of shingles costs about 8(30) = $240. These prices are close to the given values, so the solution seems reasonable.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com 6. You have a total of 18 math and science exercises for homework. You have

six more math exercises than science exercises. How many exercises do you have in each subject?

Solving Real-Life Problems

A$sr

S

1

2

3

8 16 240 32 x

80

160

240

320

0

y

(32, 20)

y = −7.5x + 260

x = 32


Tutorial Help in English and Spanish at BigIdeasMath.comExercises5.1

Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3–8, tell whether the ordered pair is a solution of the system of linear equations. (See Example 1.)

3. (2, 6); x + y = 83x − y = 0 4. (8, 2);

x − y = 62x − 10y = 4

5. (−1, 3); y = −7x − 4y = 8x + 5

6. (−4, −2); y = 2x + 6y = −3x − 14

7. (−2, 1); 6x + 5y = −72x − 4y = −8 8. (5, −6);

6x + 3y = 124x + y = 14

In Exercises 9–12, use the graph to solve the system of linear equations. Check your solution.

9. x − y = 4 10. x + y = 5 4x + y = 1 y − 2x = −4

x

y

−2

42

x

y

2

4

41

11. 6y + 3x = 18 12. 2x − y = −2 −x + 4y = 24 2x + 4y = 8

x

y

2

4

−2−4−6

x

y

4

2−2

In Exercises 13–20, solve the system of linear equations by graphing. (See Example 2.)

13. y = −x + 7 14. y = −x + 4 y = x + 1 y = 2x − 8

15. y = 1 — 3 x + 2 16. y = 3 — 4 x − 4

y = 2 — 3 x + 5 y = − 1 — 2 x + 11

17. 9x + 3y = −3 18. 4x − 4y = 20 2x − y = −4 y = −5

19. x − 4y = −4 20. 3y + 4x = 3 −3x − 4y = 12 x + 3y = −6

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in solving the system of linear equations.

21. The solution of

the linear system x − 3y = 6 and 2x − 3y = 3 is (3, −1).

✗x

y

−1

2

2

22. The solution of

the linear system y = 2x − 1 and y = x + 1 is x = 2.

✗

x

y

2

4

42

1. VOCABULARY Do the equations 5y − 2x = 18 and 6x = −4y − 10 form a system of linear equations? Explain.

2. DIFFERENT WORDS, SAME QUESTION Consider the system of linear equations −4x + 2y = 4 and 4x − y = −6. Which is different? Find “both” answers.

Solve the system of linear equations. Solve each equation for y.

Find the point of intersection of the graphs of the equations.

Find an ordered pair that is a solution of each equation in the system.

Vocabulary and Core Concept CheckVocabulary and Core Concept Check


Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencySolve the literal equation for y. (Section 1.4)

34. 10x + 5y = 5x + 20 35. 9x + 18 = 6y − 3x 36. 3 — 4 x + 1 — 4 y = 5

Reviewing what you learned in previous grades and lessons

USING TOOLS In Exercises 23–26, use a graphing calculator to solve the system of linear equations.

23. 0.2x + 0.4y = 4 24. −1.6x − 3.2y = −24 −0.6x + 0.6y = −3 2.6x + 2.6y = 26

25. −7x + 6y = 0 26. 4x − y = 1.50.5x + y = 2 2x + y = 1.5

27. MODELING WITH MATHEMATICS You have 40 minutes to exercise at the gym, and you want to burn 300 calories total using both machines. How much time should you spend on each machine? (See Example 3.)

Elliptical Trainer

8 calories per minute

Stationary Bike

6 calories per minute

28. MODELING WITH MATHEMATICS You sell small and large candlesat a craft fair. You collect $144 selling a total of 28 candles. How many of each type of candle did you sell?

29. MATHEMATICAL CONNECTIONS Write a linear equation that represents the area and a linear equation that represents the perimeter of the rectangle. Solve the system of linear equations by graphing. Interpret your solution.

6 cm

(3x − 3) cm

30. THOUGHT PROVOKING Your friend’s bank account balance (in dollars) is represented by the equation y = 25x + 250, where x is the number of months. Graph this equation. After 6 months, you want to have the same account balance as your friend. Write a linear equation that represents your account balance. Interpret the slope and y-intercept of the line that represents your account balance.

31. COMPARING METHODS Consider the equation x + 2 = 3x − 4.

a. Solve the equation using algebra.

b. Solve the system of linear equations y = x + 2 and y = 3x − 4 by graphing.

c. How is the linear system and the solution in part (b) related to the original equation and the solution in part (a)?

32. HOW DO YOU SEE IT? A teacher is purchasing binders for students. The graph shows the total costs of ordering x binders from three different companies.

1500

50

75

100

125

150

20 25 30 35

Co

st (

do

llars

)

40 45 50 x

y

Number of binders

Buying Binders

Company A

Company B

Company C

a. For what numbers of binders are the costs the same at two different companies? Explain.

b. How do your answers in part (a) relate to systems of linear equations?

33. MAKING AN ARGUMENT You and a friend are going hiking but start at different locations. You start at the trailhead and walk 5 miles per hour. Your friend starts 3 miles from the trailhead and walks 3 miles per hour.

you

your friend

a. Write and graph a system of linear equations that represents this situation.

b. Your friend says that after an hour of hiking you will both be at the same location on the trail. Is your friend correct? Use the graph from part (a) to explain your answer.

$6each $4

each

Section 5.2 Solving Systems of Linear Equations by Substitution 225

5.2 Solving Systems of Linear Equations by Substitution

Using Substitution to Solve Systems

Work with a partner. Solve each system of linear equations using two methods.

Method 1 Solve for x fi rst.Solve for x in one of the equations. Substitute the expression for x into the other equation to fi nd y. Then substitute the value of y into one of the original equations to fi nd x.

Method 2 Solve for y fi rst.Solve for y in one of the equations. Substitute the expression for y into the other equation to fi nd x. Then substitute the value of x into one of the original equations

to fi nd y.

Is the solution the same using both methods? Explain which method you would prefer

to use for each system.

a. x + y = −7 b. x − 6y = −11 c. 4x + y = −1

−5x + y = 5 3x + 2y = 7 3x − 5y = −18

Essential QuestionEssential Question How can you use substitution to solve a system of linear equations?

Writing and Solving a System of Equations

Work with a partner.

a. Write a random ordered pair with integer coordinates. One way to do this is to use a graphing calculator. The ordered pair generated at the right is (−2, −3).

b. Write a system of linear equations that has your ordered pair as its solution.

c. Exchange systems with your partner and use one of the methods from Exploration 1 to solve the system. Explain your choice of method.

Communicate Your AnswerCommunicate Your Answer 3. How can you use substitution to solve a system of linear equations?

4. Use one of the methods from Exploration 1 to solve each system of linear equations. Explain your choice of method. Check your solutions.

a. x + 2y = −7 b. x − 2y = −6 c. −3x + 2y = −10

2x − y = −9 2x + y = −2 −2x + y = −6

d. 3x + 2y = 13 e. 3x − 2y = 9 f. 3x − y = −6

x − 3y = −3 −x − 3y = 8 4x + 5y = 11

USING PRECISE MATHEMATICAL LANGUAGE

To be profi cient in math, you need to communicate precisely with others.

randInt(-5‚5‚2){-2 -3}

Choose tworandom integersbetween −5 and 5.

A.2.IA.5.C



5.2 Lesson What You Will LearnWhat You Will Learn Solve systems of linear equations by substitution.


Solving Linear Systems by SubstitutionPrevioussystem of linear equationssolution of a system of linear equations


Core Core ConceptConceptSolving a System of Linear Equations by SubstitutionStep 1 Solve one of the equations for one of the variables.

Step 2 Substitute the expression from Step 1 into the other equation and solve for the other variable.

Step 3 Substitute the value from Step 2 into one of the original equations and solve.

Solving a System of Linear Equations by Substitution

Solve the system of linear equations by substitution.

y = −2x − 9 Equation 1


SOLUTION

Step 1 Equation 1 is already solved for y.

Step 2 Substitute −2x − 9 for y in Equation 2 and solve for x.


6x − 5(−2x − 9) = −19 Substitute −2x − 9 for y.

6x + 10x + 45 = −19 Distributive Property

16x + 45 = −19 Combine like terms.

16x = −64 Subtract 45 from each side.

x = −4 Divide each side by 16.

Step 3 Substitute −4 for x in Equation 1 and solve for y.

y = −2x − 9 Equation 1

= −2(−4) − 9 Substitute −4 for x.

= 8 − 9 Multiply.

= −1 Subtract.

The solution is (−4, −1).

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comSolve the system of linear equations by substitution. Check your solution.

1. y = 3x + 14 2. 3x + 2y = 0 3. x = 6y − 7y = −4x y = 1 — 2 x − 1 4x + y = −3

Another way to solve a system of linear equations is to use substitution.

Check

Equation 1

y = −2x − 9

−1 =?

−2(−4) − 9

−1 = −1 ✓

Equation 2

6x − 5y = −19

6(−4) − 5(−1) =?

−19

−19 = −19 ✓


Solving a System of Linear Equations by Substitution

Solve the system of linear equations by substitution.

−x + y = 3 Equation 1

3x + y = −1 Equation 2

SOLUTION

Step 1 Solve for y in Equation 1.

y = x + 3 Revised Equation 1

Step 2 Substitute x + 3 for y in Equation 2 and solve for x.

3x + y = −1 Equation 2

3x + (x + 3) = −1 Substitute x + 3 for y.

4x + 3 = −1 Combine like terms.

4x = −4 Subtract 3 from each side.

x = −1 Divide each side by 4.

Step 3 Substitute −1 for x in Equation 1 and solve for y.

−x + y = 3 Equation 1

−(−1) + y = 3 Substitute −1 for x.

y = 2 Subtract 1 from each side.

The solution is (−1, 2).

Algebraic Check

Equation 1

−x + y = 3

−(−1) + 2 =?

3

3 = 3 ✓

Equation 2

3x + y = −1

3(−1) + 2 =?

−1

−1 = −1 ✓

Graphical Check

−5

−2

4

4

y = x + 3

4

y = −3x − 1

IntersectionX=-1 Y=2

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comSolve the system of linear equations by substitution. Check your solution.

4. x + y = −2 5. −x + y = −4

−3x + y = 6 4x − y = 10

6. 2x − y = −5 7. x − 2y = 7

3x − y = 1 3x − 2y = 3

ANOTHER WAYYou could also begin by solving for x in Equation 1, solving for y in Equation 2, or solving for x in Equation 2.



A drama club earns $1040 from a production. An adult ticket costs twice as much as a student ticket. Write a system of linear equations that represents this situation. What is the price of each type of ticket?

SOLUTION

1. Understand the Problem You know the amount earned, the total numbers of adult and student tickets sold, and the relationship between the price of an adult ticket and the price of a student ticket. You are asked to write a system of linear equations that represents the situation and fi nd the price of each type of ticket.

2. Make a Plan Use a verbal model to write a system of linear equations that represents the problem. Then solve the system of linear equations.


Words 64 ⋅ Adult ticket price + 132 ⋅ Student ticket price

= 1040

Adult ticket price

= 2 ⋅ Student ticket priceVariables Let x be the price (in dollars) of an adult ticket and let y be the

price (in dollars) of a student ticket.


x = 2y Equation 2

Step 1 Equation 2 is already solved for x.

Step 2 Substitute 2y for x in Equation 1 and solve for y.

64x + 132y = 1040 Equation 1

64(2y) + 132y = 1040 Substitute 2y for x.

260y = 1040 Simplify.

y = 4 Simplify.

Step 3 Substitute 4 for y in Equation 2 and solve for x.

x = 2y Equation 2

x = 2(4) Substitute 4 for y.

x = 8 Simplify.

The solution is (8, 4). So, an adult ticket costs $8 and a student ticket costs $4.

4. Look Back To check that your solution is correct, substitute the values of x and y into both of the original equations and simplify.

64(8) + 132(4) = 1040 8 = 2(4)

1040 = 1040 ✓ 8 = 8 ✓

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com 8. There are a total of 64 students in a drama club and a yearbook club. The drama

club has 10 more students than the yearbook club. Write a system of linear equations that represents this situation. How many students are in each club?


STUDY TIPYou can use either of the original equations to solve for x. However, using Equation 2 requires fewer calculations.

AtsW

S

1

2

3

Tickets sold

Type Number

adult 64

student 132



Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3−8, tell which equation you would choose to solve for one of the variables. Explain.

3. x + 4y = 30 4. 3x − y = 0 x − 2y = 0 2x + y = −10

5. 5x + 3y = 11 6. 3x − 2y = 19 5x − y = 5 x + y = 8

7. x − y = −3 8. 3x + 5y = 25 4x + 3y = −5 x − 2y = −6

In Exercises 9–16, solve the sytem of linear equations by substitution. Check your solution. (See Examples 1 and 2.)

9. x = 17 − 4y 10. 6x − 9 = y y = x − 2 y = −3x

11. x = 16 − 4y 12. −5x + 3y = 51 3x + 4y = 8 y = 10x − 8

13. 2x = 12 14. 2x − y = 23 x − 5y = −29 x − 9 = −1

15. 5x + 2y = 9 16. 11x − 7y = −14 x + y = −3 x − 2y = −4

17. ERROR ANALYSIS Describe and correct the error in solving for one of the variables in the linear system 8x + 2y = −12 and 5x − y = 4.

Step 1 5x − y = 4 −y = −5x + 4 y = 5x − 4

Step 2 5x − (5x − 4) = 4 5x − 5x + 4 = 4 4 = 4

✗

18. ERROR ANALYSIS Describe and correct the error in solving for one of the variables in the linear system 4x + 2y = 6 and 3x + y = 9.

Step 1 3x + y = 9 y = 9 − 3x

Step 2 4x + 2(9 − 3x) = 6 4x + 18 − 6x = 6 −2x = −12 x = 6

Step 3 3x + y = 9 3x + 6 = 9 3x = 3 x = 1

✗

19. MODELING WITH MATHEMATICS A farmer plants corn and wheat on a 180-acre farm. The farmer wants to plant three times as many acres of corn as wheat. Write a system of linear equations that represents this situation. How many acres of each crop should the farmer plant? (See Example 3.)

20. MODELING WITH MATHEMATICS A company that offers tubing trips down a river rents tubes for a person to use and “cooler” tubes to carry food and water. A group spends $270 to rent a total of 15 tubes. Write a system of linear equations that represents this situation. How many of each type of tube does the group rent?

1. WRITING Describe how to solve a system of linear equations by substitution.

2. NUMBER SENSE When solving a system of linear equations by substitution, how do you decide which variable to solve for in Step 1?



Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyFind the sum or difference. (Skills Review Handbook)

36. (x − 4) + (2x − 7) 37. (5y − 12) + (−5y − 1)

38. (t − 8) − (t + 15) 39. (6d + 2) − (3d − 3)

40. 4(m + 2) + 3(6m − 4) 41. 2(5v + 6) − 6(−9v + 2)


In Exercises 21–24, write a system of linear equations that has the ordered pair as its solution.

21. (3, 5) 22. (−2, 8)

23. (−4, −12) 24. (15, −25)

25. PROBLEM SOLVING A math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of each point value are on the test?

26. PROBLEM SOLVING An investor owns shares of Stock A and Stock B. The investor owns a total of 200 shares with a total value of $4000. How many shares of each stock does the investor own?

Stock Price

A $9.50

B $27.00

MATHEMATICAL CONNECTIONS In Exercises 27 and 28, (a) write an equation that represents the sum of the angle measures of the triangle and (b) use your equation and the equation shown to fi nd the values of x and y.

27.

x + 2 = 3y

x°

y°

28.

x°

y °(y − 18)°

3x − 5y = −22

29. REASONING Find the values of a and b so that the solution of the linear system is (−9, 1).

ax + by = −31 Equation 1 ax − by = −41 Equation 2

30. MAKING AN ARGUMENT Your friend says that given a linear system with an equation of a horizontal line and an equation of a vertical line, you cannot solve the system by substitution. Is your friend correct? Explain.

31. OPEN-ENDED Write a system of linear equations in which (3, −5) is a solution of Equation 1 but not a solution of Equation 2, and (−1, 7) is a solution of the system.

32. HOW DO YOU SEE IT? The graphs of two linear equations are shown.

2 4 6 x

2

4

6

y y = x + 1

y = 6 − x14

a. At what point do the lines appear to intersect?

b. Could you solve a system of linear equations by substitution to check your answer in part (a)? Explain.

33. REPEATED REASONING A radio station plays a total of 272 pop, rock, and hip-hop songs during a day. The number of pop songs is 3 times the number of rock songs. The number of hip-hop songs is 32 more than the number of rock songs. How many of each type of song does the radio station play?

34. THOUGHT PROVOKING You have $2.65 in coins. Write a system of equations that represents this situation. Use variables to represent the number of each type of coin.

35. NUMBER SENSE The sum of the digits of a two-digit number is 11. When the digits are reversed, the number increases by 27. Find the original number.

Section 5.3 Solving Systems of Linear Equations by Elimination 231

5.3 Solving Systems of Linear Equations by Elimination

Writing and Solving a System of Equations

Work with a partner. You purchase a drink and a sandwich for $4.50. Your friend purchases a drink and fi ve sandwiches for $16.50. You want to determine the price of a drink and the price of a sandwich.

a. Let x represent the price (in dollars) of one drink. Let y represent the price (in dollars) of one sandwich. Write a system of equations for the situation. Use the following verbal model.

Number of drinks ⋅ Price per drink +

Number of sandwiches ⋅ Price per sandwich = Total price

Label one of the equations Equation 1 and the other equation Equation 2.

b. Subtract Equation 1 from Equation 2. Explain how you can use the result to solve the system of equations. Then fi nd and interpret the solution.

Essential QuestionEssential Question How can you use elimination to solve a system of linear equations?

Using Elimination to Solve a System


2x + y = 7 Equation 1 x + 5y = 17 Equation 2

a. Can you eliminate a variable by adding or subtracting the equations as they are? If not, what do you need to do to one or both equations so that you can?

b. Solve the system individually. Then exchange solutions with your partner and compare and check the solutions.

Communicate Your AnswerCommunicate Your Answer 4. How can you use elimination to solve a system of linear equations?

5. When can you add or subtract the equations in a system to solve the system? When do you have to multiply fi rst? Justify your answers with examples.

6. In Exploration 3, why can you multiply an equation in the system by a constant and not change the solution of the system? Explain your reasoning.

Using Elimination to Solve Systems

Work with a partner. Solve each system of linear equations using two methods.

Method 1 Subtract. Subtract Equation 2 from Equation 1. Then use the result to solve the system.

Method 2 Add. Add the two equations. Then use the result to solve the system.

Is the solution the same using both methods? Which method do you prefer?

a. 3x − y = 6 b. 2x + y = 6 c. x − 2y = −7

3x + y = 0 2x − y = 2 x + 2y = 5

USING PROBLEM-SOLVING STRATEGIES

To be profi cient in math, you need to monitor and evaluate your progress and change course using a different solution method, if necessary.

A.2.IA.5.C



5.3 Lesson What You Will LearnWhat You Will Learn Solve systems of linear equations by elimination.


Solving Linear Systems by EliminationPreviouscoeffi cient


Core Core ConceptConceptSolving a System of Linear Equations by EliminationStep 1 Multiply, if necessary, one or both equations by a constant so at least one

pair of like terms has the same or opposite coeffi cients.

Step 2 Add or subtract the equations to eliminate one of the variables.

Step 3 Solve the resulting equation.

Step 4 Substitute the value from Step 3 into one of the original equations and solve for the other variable.

Solving a System of Linear Equations by Elimination

Solve the system of linear equations by elimination.

3x + 2y = 4 Equation 13x − 2y = −4 Equation 2

SOLUTION

Step 1 Because the coeffi cients of the y-terms are opposites, you do not need to multiply either equation by a constant.

Step 2 Add the equations.

3x + 2y = 4 Equation 13x − 2y = −4 Equation 2

6x = 0 Add the equations.

Step 3 Solve for x.

6x = 0 Resulting equation from Step 2 x = 0 Divide each side by 6.

Step 4 Substitute 0 for x in one of the original equations and solve for y.

3x + 2y = 4 Equation 13(0) + 2y = 4 Substitute 0 for x.

y = 2 Solve for y. The solution is (0, 2).

Check

Equation 1

3x + 2y = 4

3(0) + 2(2) =?

4

4 = 4 ✓Equation 2

3x − 2y = −4

3(0) − 2(2) =?

−4

−4 = −4 ✓

You can use elimination to solve a system of equations because replacing one equation in the system with the sum of that equation and a multiple of the other produces a system that has the same solution. Here is why.

Consider System 1. In this system, a and c are algebraic expressions, and b and d are constants. Begin by multiplying each side of Equation 2 by a constant k. By the Multiplication Property of Equality, kc = kd. You can rewrite Equation 1 as Equation 3 by adding kc on the left and kd on the right. You can rewrite Equation 3 as Equation 1 by subtracting kc on the left and kd on the right. Because you can rewrite either system as the other, System 1 and System 2 have the same solution.

System 1

a = b Equation 1c = d Equation 2

System 2

a + kc = b + kd Equation 3c = d Equation 2


Solving a System of Linear Equations by Elimination

Solve the system of linear equations by elimination.

−10x + 3y = 1 Equation 1

−5x − 6y = 23 Equation 2

SOLUTION

Step 1 Multiply Equation 2 by −2 so that the coeffi cients of the x-terms are opposites.

−10x + 3y = 1 −10x + 3y = 1 Equation 1

−5x − 6y = 23 Multiply by −2. 10x + 12y = −46 Revised Equation 2


− 10x + 3y = 1 Equation 1 10x + 12y = −46 Revised Equation 2

15y = −45 Add the equations.

Step 3 Solve for y.

15y = −45 Resulting equation from Step 2y = −3 Divide each side by 15.

Step 4 Substitute −3 for y in one of the original equations and solve for x.

−5x − 6y = 23 Equation 2

−5x − 6(−3) = 23 Substitute −3 for y.

−5x + 18 = 23 Multiply.

−5x = 5 Subtract 18 from each side.

x = −1 Divide each side by −5.

The solution is (−1, −3).

Monitoring Progress Help in English and Spanish at BigIdeasMath.comSolve the system of linear equations by elimination. Check your solution.

1. 3x + 2y = 7 2. x − 3y = 24 3. x + 4y = 22

−3x + 4y = 5 3x + y = 12 4x + y = 13

ANOTHER WAYTo use subtraction toeliminate one of the variables, multiply Equation 2 by 2 and then subtract the equations.

− 10x + 3y = 1−(−10x − 12y = 46) 15y = −45

Methods for Solving Systems of Linear Equations

Concept SummaryConcept Summary

Method When to Use

Graphing (Lesson 5.1) To estimate solutions

Substitution (Lesson 5.2)

When one of the variables in one of the equations has a coeffi cient of 1 or −1

Elimination (Lesson 5.3)

When at least one pair of like terms has the same or opposite coeffi cients

Elimination (Multiply First) (Lesson 5.3)

When one of the variables cannot be eliminated by adding or subtracting the equations

Check

10

−10

−10

10

IntersectionX=-1 Y=-3

10

Equation 110

3

Equation 2



The graph represents the average salaries of classroom teachers in two school districts. During what year were the average salaries in the two districts equal? What was the average salary in both districts in that year?

SOLUTION

1. Understand the Problem You know two points on each line in the graph. You are asked to determine the year in which the average salaries were equal and then determine the average salary in that year.

2. Make a Plan Use the points in the graph to write a system of linear equations. Then solve the system of linear equations.

3. Solve the Problem Find the slope of each line.

District A: m = 60 − 25 — 25 − 5

= 35 — 20

= 7 — 4 District B: m = 55 − 25 —

25 − 0 = 30 —

25 = 6 —

5

Use each slope and a point on each line to write equations of the lines.

District A District B

y − y1 = m(x − x1) Write the point-slope form. y − y1 = m(x − x1)

y − 25 = 7 — 4 (x − 5) Substitute for m, x1, and y1. y − 25 =

6 — 5 (x − 0)

−7x + 4y = 65 Write in standard form. −6x + 5y = 125

System −7x + 4y = 65 Equation 1


Step 1 Multiply Equation 1 by −5. Multiply Equation 2 by 4.

−7x + 4y = 65 Multiply by −5. 35x − 20y = −325 Revised Equation 1

−6x + 5y = 125 Multiply by 4. −24x + 20y = 500 Revised Equation 2


35x − 20y = −325 Revised Equation 1

−24x + 20y = 500 Revised Equation 2

11x = 175 Add the equations.

Step 3 Solving the equation 11x = 175 gives x = 175 — 11 ≈ 15.9.

Step 4 Substitute 175 — 11 for x in one of the original equations and solve for y.


−7 ( 175 — 11 ) + 4y = 65 Substitute 175 — 11 for x.y ≈ 44 Solve for y.

The solution is about (15.9, 44). Because x ≈ 15.9 corresponds to the year 2000, the average salary in both districts was about $44,000 in 2000.

4. Look Back Using the graph, the point of intersection appears to be about (15, 45). So, the solution of (15.9, 44) is reasonable.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com 4. Write and solve a system of linear equations represented by the graph at the left.


STUDY TIPIn Example 3, both equations are multiplied by a constant so that the coeffi cients of the y-terms are opposites.

00

20

40

10 20 30 x

y

Years since 1985

60

80

Classroom Teacher

Ave

rag

e sa

lary

(th

ou

san

ds

of

do

llars

) District ADistrict B

(25, 55)

(25, 60)

(5, 25)

(0, 25)

00

4

8

4 8 12 16 x

y

12

16

(2, 9)

(14, 15)

(4, 4)

(8, 16)Line B Line A



Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3−10, solve the system of linear equations by elimination. Check your solution. (See Example 1.)

3. x + 2y = 13 4. 9x + y = 2 −x + y = 5 −4x − y = −17

5. 5x + 6y = 50 6. −x + y = 4 x − 6y = −26 x + 3y = 4

7. −3x − 5y = −7 8. 4x − 9y = −21 −4x + 5y = 14 −4x − 3y = 9

9. −y − 10 = 6x 10. 3x − 30 = y 5x + y = −10 7y − 6 = 3x

In Exercises 11–18, solve the system of linear equations by elimination. Check your solution. (See Examples 2 and 3.)

11. x + y = 2 12. 8x − 5y = 11 2x + 7y = 9 4x − 3y = 5

13. 11x − 20y = 28 14. 10x − 9y = 46 3x + 4y = 36 −2x + 3y = 10

15. 4x − 3y = 8 16. −2x − 5y = 9 5x − 2y = −11 3x + 11y = 4

17. 9x + 2y = 39 18. 12x − 7y = −2 6x + 13y = −9 8x + 11y = 30

19. ERROR ANALYSIS Describe and correct the error in solving for one of the variables in the linear system 5x − 7y = 16 and x + 7y = 8.

5x − 7y = 16 x + 7y = 8 4x = 24 x = 6

✗

20. ERROR ANALYSIS Describe and correct the error in solving for one of the variables in the linear system 4x + 3y = 8 and x − 2y = −13.

21. MODELING WITH MATHEMATICS A service center charges a fee of x dollars for an oil change plus y dollars per quart of oil used. A sample of its sales record is shown. Write a system of linear equations that represents this situation. Find the fee and cost per quart of oil.

A B

2

1

34

Customer Oil Tank Size(quarts)TotalCost

A 5B 7

C

$22.45$25.45

22. MODELING WITH MATHEMATICS The graph represents the average salaries of high school principals in two states. During what year were the average salaries in the two states equal? What was the average salary in both states in that year?

1. OPEN-ENDED Give an example of a system of linear equations that can be solved by fi rst adding the equations to eliminate one variable.

2. WRITING Explain how to solve the system of linear equations 2x − 3y = −4 Equation 1−5x + 9y = 7 Equation 2by elimination.


4x + 3y = 8 4x + 3y = 8

x − 2y = −13 Multiply by −4. −4x + 8y = −13

11y = −5

y = −5 — 11

✗

00

40

80

10 20 30 x

y

Years since 1985

120

160

High School Principal

Ave

rag

e sa

lary

(th

ou

san

ds

of

do

llars

)

State AState B

(22, 100)

(25, 98)(0, 43)

(6, 48)


Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencySolve the equation. Determine whether the equation has one solution, no solution, or infi nitely many solutions. (Section 1.3)

36. 5d − 8 = 1 + 5d 37. 9 + 4t = 12 − 4t

38. 3n + 2 = 2(n − 3) 39. −3(4 − 2v) = 6v − 12

Write an equation of the line that passes through the given point and is parallel to the given line. (Section 4.4)

40. (4, −1); y = −2x + 7 41. (0, 6); y = 5x − 3 42. (−5, −2); y = 2 — 3 x + 1


In Exercises 23–26, solve the system of linear equations using any method. Explain why you chose the method.

23. 3x + 2y = 4 24. −6y + 2 = −4x 2y = 8 − 5x y − 2 = x

25. y − x = 2 26. 3x + y = 1 — 3

y = − 1 — 4 x + 7 2x − 3y = 8 — 3

27. WRITING For what values of a can you solve the linear system ax + 3y = 2 and 4x + 5y = 6 by elimination without multiplying fi rst? Explain.

28. HOW DO YOU SEE IT? The circle graph shows the results of a survey in which 50 students were asked about their favorite meal.

Favorite Meal

Dinner25

Breakfast

Lunch

a. Estimate the numbers of students who chose breakfast and lunch.

b. The number of students who chose lunch was 5 more than the number of students who chose breakfast. Write a system of linear equations that represents the numbers of students who chose breakfast and lunch.

c. Explain how you can solve the linear system in part (b) to check your answers in part (a).

29. MAKING AN ARGUMENT Your friend says that any system of equations that can be solved by elimination can be solved by substitution in an equal or fewer number of steps. Is your friend correct? Explain.

30. THOUGHT PROVOKING Write a system of linear equations that can be added to eliminate a variable or subtracted to eliminate a variable.

31. MATHEMATICAL CONNECTIONS A rectangle has a perimeter of 18 inches. A new rectangle is formed by doubling the width w and tripling the lengthℓ, as shown. The new rectangle has a perimeter P of 46 inches.

P = 46 in. 2w

3

a. Write and solve a system of linear equations to fi nd the length and width of the original rectangle.

b. Find the length and width of the new rectangle.

32. CRITICAL THINKING Refer to the discussion of System 1 and System 2 on page 232. Without solving, explain why the two systems shown have the same solution.

System 1 System 2

3x − 2y = 8 Equation 1 5x = 20 Equation 3x + y = 6 Equation 2 x + y = 6 Equation 2

33. PROBLEM SOLVING You are making 6 quarts of fruit punch for a party. You have bottles of 100% fruit juice and 20% fruit juice. How many quarts of each type of juice should you mix to make 6 quarts of 80% fruit juice?

34. PROBLEM SOLVING A motorboat takes 40 minutes to travel 20 miles downstream. The return trip takes 60 minutes. What is the speed of the current?

35. CRITICAL THINKING Solve for x, y, and z in the system of equations. Explain your steps.

x + 7y + 3z = 29 Equation 1 3z + x − 2y = −7 Equation 2 5y = 10 − 2x Equation 3

Section 5.4 Solving Special Systems of Linear Equations 237

5.4 Solving Special Systems of Linear Equations

Using a Table to Solve a System

Work with a partner. You invest $450 for equipment to make skateboards. The materials for each skateboard cost $20. You sell each skateboard for $20.

a. Write the cost and revenue equations. Then copy and complete the table for your cost C and your revenue R.

b. When will your company break even? What is wrong?

Essential QuestionEssential Question Can a system of linear equations have no solution or infi nitely many solutions?

Writing and Analyzing a System

Work with a partner. A necklace and matching bracelet have two types of beads. The necklace has 40 small beads and 6 large beads and weighs 10 grams. The bracelet has 20 small beads and 3 large beads and weighs 5 grams. The threads holding the beads have no signifi cant weight.

a. Write a system of linear equations that represents the situation. Let x be the weight (in grams) of a small bead and let y be the weight (in grams) of a large bead.

b. Graph the system in the coordinate plane shown. What do you notice about the two lines?

c. Can you fi nd the weight of each type of bead? Explain your reasoning.

Communicate Your AnswerCommunicate Your Answer 3. Can a system of linear equations have no solution or infi nitely many solutions?

Give examples to support your answers.

4. Does the system of linear equations represented by each graph have no solution, one solution, or infi nitely many solutions? Explain.

a.

x

y

4

1

42−1

y = x + 2

x + y = 2

b.

x

y

3

6

42

3

y = x + 2

−x + y = 1

c.

x

y

3

1

6

42

3

y = x + 2

−2x + 2y = 4

x (skateboards) 0 1 2 3 4 5 6 7 8 9 10

C (dollars)

R (dollars)

APPLYING MATHEMATICSTo be profi cient in math, you need to interpret mathematical results in real-life contexts.

x

y

1

1.5

2

0.5

00.2 0.3 0.40.10

A.2.IA.3.FA.5.C



5.4 Lesson What You Will LearnWhat You Will Learn Determine the numbers of solutions of linear systems.

Use linear systems to solve real-life problems.

The Numbers of Solutions of Linear SystemsPreviousparallel


Core Core ConceptConceptSolutions of Systems of Linear EquationsA system of linear equations can have one solution, no solution, or infi nitely many solutions.

One solution No solution Infi nitely many solutions

x

y

x

y

x

y

The lines intersect. The lines are parallel. The lines are the same.

Solving a System: No Solution

Solve the system of linear equations.

y = 2x + 1 Equation 1y = 2x − 5 Equation 2

SOLUTION

Method 1 Solve by graphing.

Graph each equation.

The lines have the same slope and different y-intercepts. So, the lines are parallel.

Because parallel lines do not intersect, there is no point that is a solution of both equations.

So, the system of linear equationshas no solution.

Method 2 Solve by substitution.

Substitute 2x − 5 for y in Equation 1.

y = 2x + 1 Equation 1

2x − 5 = 2x + 1 Substitute 2x − 5 for y.

−5 = 1 ✗ Subtract 2x from each side. The equation −5 = 1 is never true. So, the system of linear equations

has no solution.

ANOTHER WAYYou can solve some linear systems by inspection. In Example 1, notice you can rewrite the system as

–2x + y = 1–2x + y = –5.

This system has no solution because –2x + y cannot be equal to both 1 and –5.

STUDY TIPA linear system with no solution is called an inconsistent system.

x

y

2

−2

−4

41−21

2

1

2

2

−22

y = 2x + 1

x44

y = 2x − 5


Solving a System: Infi nitely Many Solutions

Solve the system of linear equations.

−2x + y = 3 Equation 1


SOLUTION

Solve by elimination.

Step 1 Multiply Equation 1 by −2.

−2x + y = 3 Multiply by −2. 4x − 2y = −6 Revised Equation 1

−4x + 2y = 6 −4x + 2y = 6 Equation 2


4x − 2y = −6 Revised Equation 1−4x + 2y = 6 Equation 2

0 = 0 Add the equations.

The equation 0 = 0 is always true. So, the solutions are all the points on the line −2x + y = 3. The system of linear equations has infi nitely many solutions.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comSolve the system of linear equations.

1. x + y = 3 2. y = −x + 3

2x + 2y = 6 2x + 2y = 4

ANOTHER WAYYou can also solve the linear system by graphing. The lines have the same slope and the same y-intercept. So, the lines are the same, which means all points on the line are solutions of both equations.

STUDY TIPA linear system with infi nitely many solutions is called a consistent dependent system.



An athletic director is comparing the costs of renting two banquet halls for an awards banquet. Write a system of linear equations that represents this situation. If the cost patterns continue, will the cost of Hall A ever equal the cost of Hall B?

SOLUTION

Words Total cost = Cost per hour ⋅ Number of hours + Initial costVariables Let y be the cost (in dollars) and let x be the number of hours.

System y = 100x + 75 Equation 1 - Cost of Hall A

y = 100x + 100 Equation 2 - Cost of Hall B

The equations are in slope-intercept form. The graphs of the equations have the same slope but different y-intercepts. There is no solution because the lines are parallel.

So, the cost of Hall A will never equal the cost of Hall B.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com 3. WHAT IF? What happens to the solution in Example 3 when the cost per hour for

Hall A is $125?

Cost (dollars)

Hours Hall A Hall B

0 75 100

1 175 200

2 275 300

3 375 400



The perimeter of the trapezoidal piece of land is 48 kilometers. The perimeter of the rectangular piece of land is 144 kilometers. Write and solve a system of linear equations to fi nd the values of x and y.

SOLUTION

1. Understand the Problem You know the perimeter of each piece of land and the side lengths in terms of x or y. You are asked to write and solve a system of linear equations to fi nd the values of x and y.

2. Make a Plan Use the fi gures and the defi nition of perimeter to write a system of linear equations that represents the problem. Then solve the system of linear equations.


Perimeter of trapezoid Perimeter of rectangle

2x + 4x + 6y + 6y = 48 9x + 9x + 18y + 18y = 144

6x + 12y = 48 Equation 1 18x + 36y = 144 Equation 2


18x + 36y = 144 Equation 2

Method 1 Solve by graphing.

Graph each equation.

The lines have the same slope and the same y-intercept. So, the lines are the same.

In this context, x and y must be positive. Because the lines are the same, all the points on the line in Quadrant I are solutions of both equations.

So, the system of linear equations has infi nitely many solutions.

Method 2 Solve by elimination.

Multiply Equation 1 by −3 and add the equations.

6x + 12y = 48 Multiply by −3. −18x − 36y = −144 Revised Equation 1

18x + 36y = 144 18x + 36y = 144 Equation 2

0 = 0 Add the equations.

The equation 0 = 0 is always true. In this context, x and y must be positive. So, the solutions are all the points on the line 6x + 12y = 48 in Quadrant I. The system of linear equations has infi nitely many solutions.

4. Look Back Choose a few of the ordered pairs (x, y) that are solutions of Equation 1. You should fi nd that no matter which ordered pairs you choose, they will also be solutions of Equation 2. So, infi nitely many solutions seems reasonable.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com 4. WHAT IF? What happens to the solution in Example 4 when the perimeter of the

trapezoidal piece of land is 96 kilometers? Explain.

4x

2x

6y 6y

18y

18y

9x9x

2 4 60 x

2

4

6

0

y

6x + 12y = 486x + 12y =

18x + 36y = 144



Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3−8, match the system of linear equations with its graph. Then determine whether the system has one solution, no solution, or infi nitely many solutions.

3. −x + y = 1 4. 2x − 2y = 4 x − y = 1 −x + y = −2

5. 2x + y = 4 6. x − y = 0 −4x − 2y = −8 5x − 2y = 6

7. −2x + 4y = 1 8. 5x + 3y = 17 3x − 6y = 9 x − 3y = −2

A.

x

y

2

4

2−1

B.

x

y

2

4

6

1 4−2

C.

x

y

2

−2

2 4−2

D.

x

y

2

−3

1 4

E.

x

y

2

−1 2 4

F.

x

y

2

−3

2−3

In Exercises 9–16, solve the system of linear equations. (See Examples 1 and 2.)

9. y = −2x − 4 10. y = −6x − 8 y = 2x − 4 y = −6x + 8

11. 3x − y = 6 12. −x + 2y = 7 −3x + y = −6 x − 2y = 7

13. 4x + 4y = −8 14. 15x − 5y = −20 −2x − 2y = 4 −3x + y = 4

15. 9x − 15y = 24 16. 3x − 2y = −5 6x − 10y = −16 4x + 5y = 47

In Exercises 17–22, use only the slopes and y-intercepts of the graphs of the equations to determine whether the system of linear equations has one solution, no solution, or infi nitely many solutions. Explain.

17. y = 7x + 13 18. y = −6x − 2 −21x + 3y = 39 12x + 2y = −6

19. 4x + 3y = 27 20. −7x + 7y = 1 4x − 3y = −27 2x − 2y = −18

21. −18x + 6y = 24 22. 2x − 2y = 16 3x − y = −2 3x − 6y = 30

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in solving the system of linear equations.

23. −4x + y = 4 4x + y = 12

The lines do not intersect. So, the system has no solution.

✗x

y

1

−3

2−2

24. y = 3x − 8 y = 3x − 12

The lines have the same slope. So, the system has infi nitely many solutions.

✗

1. REASONING Is it possible for a system of linear equations to have exactly two solutions? Explain.

2. WRITING Compare the graph of a system of linear equations that has infi nitely many solutions and the graph of a system of linear equations that has no solution.



Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencySolve the system of linear equations by graphing. (Section 5.1)

33. y = x − 6 34. y = 2x + 3 35. 2x + y = 6y = −x + 10 y = −3x − 7 3x − 2y = 16


25. MODELING WITH MATHEMATICS The table shows the distances two groups have traveled at different times during a canoeing excursion. The groups continue traveling at their current rates until they reach the same destination. Let d be the distance traveled and t be the time since 1 p.m. Write a system of linear equations that represents this situation. Will Group B catch up to Group A before reaching the destination? Explain. (See Example 3.)

Distance Traveled (miles)

1 P.M. 2 P.M. 3 P.M. 4 P.M.

Group A 3 9 15 21

Group B 1 7 13 19

26. MODELING WITH MATHEMATICS A $6-bag of trail mix contains 3 cups of dried fruit and 4 cups of almonds. A $9-bag contains 4 1 — 2 cups of dried fruit and 6 cups of almonds. Write and solve a system of linear equations to fi nd the price of 1 cup of dried fruit and 1 cup of almonds. (See Example 4.)

27. PROBLEM SOLVING A train travels from New York City to Washington, D.C., and then back to New York City. The table shows the number of tickets purchased for each leg of the trip. The cost per ticket is the same for each leg of the trip. Is there enough information to determine the cost of one coach ticket? Explain.

DestinationCoach tickets

Business class

tickets

Money collected (dollars)

Washington, D.C. 150 80 22,860

New York City 170 100 27,280

28. THOUGHT PROVOKING Write a system of three linear equations in two variables so that any two of the equations have exactly one solution, but the entire system of equations has no solution.

29. REASONING In a system of linear equations, one equation has a slope of 2 and the other equation has a slope of − 1 — 3 . How many solutions does the system have? Explain.

30. HOW DO YOU SEE IT? The graph shows information about the last leg of a 4 × 200-meter relay for three relay teams. Team A’s runner ran about 7.8 meters per second, Team B’s runner ran about 7.8 meters per second, and Team C’s runner ran about 8.8 meters per second.

400

50

100

150

8 12 16 20 24 28

Dis

tan

ce (

met

ers)

x

y

Time (seconds)

Last Leg of 4 × 200-Meter Relay

Team CTeam B

Team A

a. Estimate the distance at which Team C’s runner passed Team B’s runner.

b. If the race was longer, could Team C’s runner have passed Team A’s runner? Explain.

c. If the race was longer, could Team B’s runner have passed Team A’s runner? Explain.

31. ABSTRACT REASONING Consider the system of linear equations y = ax + 4 and y = bx − 2, where a and b are real numbers. Determine whether each statement is always, sometimes, or never true. Explain your reasoning.

a. The system has infi nitely many solutions.

b. The system has no solution.

c. When a < b, the system has one solution.

32. MAKING AN ARGUMENT One admission to an ice skating rink costs x dollars, and renting a pair of ice skates costs y dollars. Your friend says she can determine the exact cost of one admission and one skate rental. Is your friend correct? Explain.

Total

2Admissions3

38.00

Skate Rentals

$ Total

15 Admissions10

190.00

Skate Rentals

$

243243

5.1–5.4 What Did You Learn?

Study Errors

What Happens: You do not study the right material or you do not learn it well enough to remember it on a test without resources such as notes.

How to Avoid This Error: Take a practice test. Work with a study group. Discuss the topics on the test with your teacher. Do not try to learn a whole chapter’s worth of material in one night.

Core VocabularyCore Vocabularysystem of linear equations, p. 220 solution of a system of linear equations, p. 220

Core ConceptsCore ConceptsSection 5.1Solving a System of Linear Equations by Graphing, p. 221

Section 5.2Solving a System of Linear Equations by Substitution, p. 226

Section 5.3Solving a System of Linear Equations by Elimination, p. 232

Section 5.4Solutions of Systems of Linear Equations, p. 238

Mathematical ThinkingMathematical Thinking1. Describe the given information in Exercise 33 on page 230 and your plan for fi nding the solution.

2. Describe another real-life situation similar to Exercise 21 on page 235 and the mathematics that you can apply to solve the problem.

3. What question(s) can you ask your friend to help her understand the error in the statement she made in Exercise 32 on page 242?

Study Skills

Analyzing Your Errors


5.1–5.4 Quiz

Use the graph to solve the system of linear equations. Check your solution. (Section 5.1)

1. y = − 1 — 3 x + 2 2. y = 1 — 2 x − 1 3. y = 1

y = x − 2 y = 4x + 6 y = 2x + 1

x

y3

1

−142−2

x

y2

−2

−4

−2−4

x

y

2

−2

2−2

Solve the system of linear equations by substitution. Check your solution. (Section 5.2)

4. y = x − 4 5. 2y + x = −4 6. 3x − 5y = 13−2x + y = 18 y − x = −5 x + 4y = 10

Solve the system of linear equations by elimination. Check your solution. (Section 5.3)

7. x + y = 4 8. x + 3y = 1 9. 2x − 3y = −5−3x − y = −8 5x + 6y = 14 5x + 2y = 16

Solve the system of linear equations. (Section 5.4)

10. x − y = 1 11. 6x + 2y = 16 12. 3x − 3y = −2x − y = 6 2x − y = 2 −6x + 6y = 4

13. You plant a spruce tree that grows 4 inches per year and a hemlock tree that grows 6 inches per year. The initial heights are shown. (Section 5.1)

a. Write a system of linear equations that represents this situation.

b. Solve the system by graphing. Interpret your solution.

14. It takes you 3 hours to drive to a concert 135 miles away. You drive 55 miles per hour on highways and 40 miles per hour on the rest of the roads. (Section 5.1, Section 5.2, and Section 5.3)

a. How much time do you spend driving at each speed?

b. How many miles do you drive on highways? the rest of the roads?

15. In a football game, all of the home team’s points are from 7-point touchdowns and 3-point fi eld goals. The team scores six times. Write and solve a system of linear equations to fi nd the numbers of touchdowns and fi eld goals that the home team scores. (Section 5.1, Section 5.2, and Section 5.3)

sprucetree

hemlocktree

14 in.

8 in.

Section 5.5 Solving Equations by Graphing 245

Solving Equations by Graphing5.5

Solving an Equation by Graphing

Work with a partner. Solve 2x − 1 = − 1 — 2 x + 4 by graphing.

a. Use the left side to write a linear equation. Then use the right side to write another linear equation.

b. Graph the two linear equations from part (a). Find the x-value of the point of intersection. Check that the x-value is the solution of

2x − 1 = − 1 — 2 x + 4.

c. Explain why this “graphical method” works.

Essential QuestionEssential Question How can you use a system of linear equations to solve an equation with variables on both sides?

Previously, you learned how to use algebra to solve equations with variables on both sides. Another way is to use a system of linear equations.

Solving Equations Algebraically and Graphically

Work with a partner. Solve each equation using two methods.

Method 1 Use an algebraic method.

Method 2 Use a graphical method.

Is the solution the same using both methods?

a. 1 — 2 x + 4 = − 1 — 4 x + 1 b.

2 — 3 x + 4 =

1 — 3 x + 3

c. − 2 — 3 x − 1 = 1 — 3 x − 4 d.

4 — 5 x +

7 — 5 = 3x − 3

e. −x + 2.5 = 2x − 0.5 f. − 3x + 1.5 = x + 1.5

Communicate Your AnswerCommunicate Your Answer 3. How can you use a system of linear equations to solve an equation with

variables on both sides?

4. Compare the algebraic method and the graphical method for solving a linear equation with variables on both sides. Describe the advantages and disadvantages of each method.

SELECTING TOOLSTo be profi cient in math, you need to consider the available tools, which may include pencil and paper or a graphing calculator, when solving a mathematical problem.

x

y

2

4

6

2 4 6−2

−2

A.5.A



5.5 Lesson What You Will LearnWhat You Will Learn Solve linear equations by graphing. Use linear equations to solve real-life problems.

Solving Linear Equations by GraphingYou can use a system of linear equations to solve an equation with variables on both sides.

Previousabsolute value equation


Core Core ConceptConceptSolving Linear Equations by GraphingStep 1 To solve the equation ax + b = cx + d, write two linear equations.

ax + b = cx + d

and

Step 2 Graph the system of linear equations. The x-value of the solution of the system of linear equations is the solution of the equation ax + b = cx + d.

y = cx + dy = ax + b

Solving an Equation by Graphing

Solve −x + 1 = 2x − 5 by graphing. Check your solution.

SOLUTION

Step 1 Write a system of linear equations using each side of the original equation.

−x + 1 = 2x − 5

Step 2 Graph the system.

y = −x + 1 Equation 1 y = 2x − 5 Equation 2

The graphs intersect at (2, −1).

So, the solution of the equation is x = 2.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comSolve the equation by graphing. Check your solution.

1. 1 — 2 x − 3 = 2x 2. −4 + 9x = −3x + 2

Check

−x + 1 = 2x − 5

−(2) + 1 =?

2(2) − 5

−1 = −1 ✓

y = 2x − 5y = −x + 1

x

y1

1

y = −x + 1

y = 2x − 5

−1

−2

−4

(2, −1)




You are an ecologist studying the populations of two types of fi sh in a lake. Use the information in the table to predict when the populations of the two types of fi sh will be equal.

SOLUTION

1. Understand the Problem You know the current population of each type of fi sh and the annual change in each population. You are asked to determine when the populations of the two types of fi sh will be equal.

2. Make a Plan Use a verbal model to write an equation that represents the problem. Then solve the equation by graphing.


Words Type A Type B

Change per year ⋅ Years +

Currentpopulation

= Change per year

⋅ Years + Currentpopulation Variable Let x be the number of years.

Equation −250x + 5750 = 400x + 2825

Solve the equation by graphing.


−250x + 5750 = 400x + 2825

Step 2 Graph the system.

y = −250x + 5750 Equation 1

y = 400x + 2825 Equation 2

The graphs appear to intersect at (4.5, 4625). Check this solution in each equation of the linear system, as shown.

So, the populations of the two types of fi sh will be equal in 4.5 years.

4. Look Back To check that your solution is correct, verify that x = 4.5 is the solution of the original equation.

−250(4.5) + 5750 = 400(4.5) + 2825

4625 = 4625 ✓

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com 3. WHAT IF? Type C has a current population of 3500 and grows by 500 each year.

Predict when the population of Type C will be equal to the population of Type A.

Check

y = −250x + 5750

4625 =?

−250(4.5) + 5750

4625 = 4625 ✓

y = 400x + 2825

4625 =?

400(4.5) + 2825

4625 = 4625 ✓Years

Fish

po

pu

lati

on

x

y

2500

3500

4500

5500

6500

02 3 4 510

(0, 5750)

(4.5, 4625)

(0, 2825)

750)

(4.5, 4625)

y = −250x + 5750

))y = 400x + 2825

Fish in a Lake

TypeCurrent

populationChange per year

A 5750 −250

B 2825 400

y = 400x + 2825y = −250x + 5750



Your family needs to rent a car for a week while on vacation. Company A charges $3.25 per mile plus a fl at fee of $125 per week. Company B charges $3 per mile plus a fl at fee of $150 per week. After how many miles of travel are the total costs the same at both companies?

SOLUTION

1. Understand the Problem You know the costs of renting a car from two companies. You are asked to determine how many miles of travel will result in the same total costs at both companies.

2. Make a Plan Use a verbal model to write an equation that represents the problem. Then solve the equation by graphing.


Words Company A Company B

Cost per mile ⋅ Miles +

Flatfee

= Cost per mile

⋅ Miles + FlatfeeVariable Let x be the number of miles traveled.

Equation 3.25x + 125 = 3x + 150

Solve the equation by graphing.


3.25x + 125 = 3x + 150

Step 2 Use a graphing calculator to graph the system.

00

600

150IntersectionX=100 Y=450

ction

y = 3.25x + 125

y = 3x + 150

Because the graphs intersect at (100, 450), the solution of the equation is x = 100.

So, the total costs are the same after 100 miles.

4. Look Back One way to check your solution is to solve the equation algebraically, as shown.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.com 4. WHAT IF? Company C charges $3.30 per mile plus a fl at fee of $115 per

week. After how many miles are the total costs the same at Company A and Company C?

y = 3x + 150y = 3.25x + 125

Check

3.25x + 125 = 3x + 150 0.25x + 125 = 150 0.25x = 25 x = 100



Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with MathematicsIn Exercises 3–10, use the graph to solve the equation. Check your solution.

3. x − 3 = 2 4. −3 = 4x + 1

x

y

−2

1

3

1 4

x

y

1

2−2

5. −2x + 3 = x 6. x − 4 = 3x

x

y

1

3

1 3−3

xy

−6

2−2

7. 2x + 1 = −x − 2 8. −5x − 1 = 3x − 1

x

y

−3

2

2−2

x

y

−1 2−2

9. −x − 1 = 1 — 3 x + 3 10. − 3 — 2 x − 2 = −4x + 3

x

y

2

4

1−2−4

x

y

2 4

−2

−4

−6

In Exercises 11−18, solve the equation by graphing. Check your solution. (See Example 1.)

11. x + 4 = −x 12. 4x = x + 3

13. x + 5 = −2x − 4 14. −2x + 6 = 5x − 1

15. 1 — 2 x − 2 = 9 − 5x 16. −5 + 1 — 4 x = 3x + 6

17. 5x − 7 = 2(x + 1) 18. −6(x + 4) = −3x − 6

In Exercises 19−24, solve the equation by graphing. Determine whether the equation has one solution, no solution, or infi nitely many solutions.

19. 3x − 1 = −x + 7 20. 5x − 4 = 5x + 1

21. −4(2 − x) = 4x − 8

22. −2x − 3 = 2(x − 2)

23. −x − 5 = − 1 — 3 (3x + 5)

24. 1 — 2 (8x + 3) = 4x + 3 — 2

In Exercises 25–28, write an equation that has the same solution as the linear system represented by the graph.

25.

x

y

−2

2

4

2 4

26.

x

y

2

4

2 4

27.

x

y

−4

2

2−4

28.

x

y

2

−2

4

1. REASONING The graphs of the equations y = 3x − 20 and y = −2x + 10 intersect at the point (6, −2). Without solving, fi nd the solution of the equation 3x − 20 = −2x + 10.

2. WRITING Consider the equation ax + b = cx + d, where c = 0. Can you solve the equation by graphing a system of linear equations? Explain.



Maintaining Mathematical ProficiencyMaintaining Mathematical ProficiencyGraph the inequality. (Section 2.1)

40. y > 5 41. x ≤ −2 42. n ≥ 9 43. c < −6

Use the graphs of f and g to describe the transformation from the graph of f to the graph of g. (Section 3.7)

44. f(x) = x − 5; g(x) = f(x + 2) 45. f(x) = 6x; g(x) = −f(x)

46. f(x) = −2x + 1; g(x) = f(4x) 47. f(x) = 1 — 2 x − 2; g(x) = f(x − 1)


USING TOOLS In Exercises 29 and 30, use a graphing calculator to solve the equation.

29. 0.7x + 0.5 = −0.2x − 1.3

30. 2.1x + 0.6 = −1.4x + 6.9

31. MODELING WITH MATHEMATICS You need to hire a catering company to serve meals to guests at a wedding reception. Company A charges $500 plus $20 per guest. Company B charges $800 plus $16 per guest. For how many guests are the total costs the same at both companies? (See Examples 2 and 3.)

32. MODELING WITH MATHEMATICS Your dog is 16 years old in dog years. Your cat is 28 years old in cat years. For every human year, your dog ages by 7 dog years and your cat ages by 4 cat years. In how many human years will both pets be the same age in their respective types of years?

33. MODELING WITH MATHEMATICS You and a friend race across a fi eld to a fence. Your friend has a 50-meter head start. The equations shown represent you and your friend’s distances d (in meters) from the fence t seconds after the race begins. Find the time at which you catch up to your friend.

You: d = −5t + 200

Your friend: d = −3 1 — 3 t + 150

34. MAKING AN ARGUMENT The graphs of y = −x + 4 and y = 2x − 8 intersect at the point (4, 0). So, your friend says the solution of the equation −x + 4 = 2x − 8 is (4, 0). Is your friend correct? Explain.

35. OPEN-ENDED Find values for m and b so that the solution of the equation mx + b = − 2x − 1 is x = −3.

36. HOW DO YOU SEE IT? The graph shows the total revenue and expenses of a company x years after it opens for business.

200

2

4

6

4 6 8 10

Mill

ion

s o

f d

olla

rs

x

y

Year

Revenue and Expenses

revenue

expenses

a. Estimate the point of intersection of the graphs.

b. Interpret your answer in part (a).

37. MATHEMATICAL CONNECTIONS The value of the perimeter of the triangle (in feet) is equal to the value of the area of the triangle (in square feet). Use a graph to fi nd x.

38. THOUGHT PROVOKING A car has an initial value of $20,000 and decreases in value at a rate of $1500 per year. Describe a different car that will be worth the same amount as this car in exactly 5 years. Specify the initial value and the rate at which the value decreases.

39. ABSTRACT REASONING Use a graph to determine the sign of the solution of the equation ax + b = cx + d in each situation.

a. 0 < b < d and a < c b. d < b < 0 and a < c

6 ftx ft

(x − 2) ft

Section 5.6 Linear Inequalities in Two Variables 251

5.6 Linear Inequalities in Two Variables

Essential QuestionEssential Question How can you write and graph a linear inequality in two variables?

A solution of a linear inequality in two variables is an ordered pair (x, y) that makes the inequality true. The graph of a linear inequality in two variables shows all the solutions of the inequality in a coordinate plane.

Writing a Linear Inequality in Two Variables


a. Write an equation represented by the dashed line.

b. The solutions of an inequality are represented by the shaded region. In words, describe the solutions of the inequality.

c. Write an inequality represented by the graph. Which inequality symbol did you use? Explain your reasoning.

Graphing Linear Inequalities in Two Variables

Work with a partner. Graph each linear inequality in two variables. Explain your steps. Use a graphing calculator to check your graphs.

a. y > x + 5 b. y ≤ − 1 — 2 x + 1 c. y ≥ −x − 5

Communicate Your AnswerCommunicate Your Answer 4. How can you write and graph a linear inequality in two variables?

5. Give an example of a real-life situation that can be modeled using a linear inequality in two variables.

SELECTING TOOLSTo be profi cient in math, you need to use technological tools to explore and deepen your understanding of concepts.

x

y

2

4

2 4−2−4

−2


Work with a partner. Use a graphing calculator to graph y ≥ 1 — 4 x − 3.

a. Enter the equation y = 1 — 4 x − 3 into your calculator.

b. The inequality has the symbol ≥. So, the region to be shaded is above the graph of y = 1 — 4 x − 3, as shown. Verify this by testing a point in this region, such as (0, 0), to make sure it is a solution of the inequality.

Because the inequality symbol is greater than or equal to, the line is solid and not dashed. Some graphing calculators always use a solid line when graphing inequalities. In this case, you have to determine whether the line should be solid or dashed, based on the inequality symbol used in the original inequality.

10

−10

−10

10

y ≥ x − 314

A.2.HA.3.D



5.6 Lesson What You Will LearnWhat You Will Learn Check solutions of linear inequalities. Graph linear inequalities in two variables.

Write linear inequalities in two variables.

Use linear inequalities to solve real-life problems.

Linear InequalitiesA linear inequality in two variables, x and y, can be written as

ax + by < c ax + by ≤ c ax + by > c ax + by ≥ c

where a, b, and c are real numbers. A solution of a linear inequality in two variables is an ordered pair (x, y) that makes the inequality true.

linear inequality in two variables, p. 252solution of a linear inequality in two variables, p. 252graph of a linear inequality, p. 252half-planes, p. 252

Previousordered pair


Checking Solutions

Tell whether the ordered pair is a solution of the inequality.

a. 2x + y < −3; (−1, 9) b. x − 3y ≥ 8; (2, −2)

SOLUTION

a. 2x + y < −3 Write the inequality.

2(−1) + 9 <?

−3 Substitute −1 for x and 9 for y.

7 < −3 ✗ Simplify. 7 is not less than −3. So, (−1, 9) is not a solution of the inequality.

b. x − 3y ≥ 8 Write the inequality.

2 − 3(−2) ≥?

8 Substitute 2 for x and −2 for y.

8 ≥ 8 ✓ Simplify. 8 is equal to 8. So, (2, −2) is a solution of the inequality.

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comTell whether the ordered pair is a solution of the inequality.

1. x + y > 0; (−2, 2) 2. 4x − y ≥ 5; (0, 0)

3. 5x − 2y ≤ −1; (−4, −1) 4. −2x − 3y < 15; (5, −7)

Graphing Linear Inequalities in Two VariablesThe graph of a linear inequality in two variables shows all the solutions of the inequality in a coordinate plane.

x

y4

2

2−2

All solutions of < 2lie on one side of the = 2 .

The boundary line dividesthe coordinate plane into twohalf-planes. The shadedhalf-plane is the graph of < 2 .

y

y

x

xy x

boundary line

READINGA dashed boundary line means that points on the line are not solutions. A solid boundary line means that points on the line are solutions.

Section 5.6 Linear Inequalities in Two Variables 253

Graphing a Linear Inequality in One Variable

Graph y ≤ 2 in a coordinate plane.

SOLUTION

Step 1 Graph y = 2. Use a solid line because the

x

y

1

3

2 4−1

(0, 0)

inequality symbol is ≤.

Step 2 Test (0, 0).

y ≤ 2 Write the inequality.

0 ≤ 2 ✓ Substitute.Step 3 Because (0, 0) is a solution, shade the

half-plane that contains (0, 0).

Check

3

−1

−2

5

Core Core ConceptConceptGraphing a Linear Inequality in Two VariablesStep 1 Graph the boundary line for the inequality. Use a dashed line for < or >.

Use a solid line for ≤ or ≥.

Step 2 Test a point that is not on the boundary line to determine whether it is a solution of the inequality.

Step 3 When the test point is a solution, shade the half-plane that contains the point. When the test point is not a solution, shade the half-plane that does not contain the point.

Graphing a Linear Inequality in Two Variables

Graph −x + 2y > 2 in a coordinate plane.

SOLUTION

Step 1 Graph −x + 2y = 2, or y = 1 — 2 x + 1. Use a

x

y

2

4

2−2

(0, 0)

dashed line because the inequality symbol is >.

Step 2 Test (0, 0).

−x + 2y > 2 Write the inequality.

−(0) + 2(0) >?

2 Substitute.

0 > 2 ✗ Simplify.Step 3 Because (0, 0) is not a solution, shade the

half-plane that does not contain (0, 0).

Monitoring ProgressMonitoring Progress Help in English and Spanish at BigIdeasMath.comGraph the inequality in a coordinate plane.

5. y > −1 6. x ≤ −4

7. x + y ≤ −4 8. x − 2y < 0

STUDY TIPIt is often convenient to use the origin as a test point. However, you must choose a different test point when the origin is on the boundary line.


Writing Linear Inequalities in Two Variables

Core Core ConceptConceptWriting a Linear Inequality in Two Variables Using a GraphWrite an equation in slope-intercept form of the boundary line.

If the shaded half-plane is above the boundary line, then

• replace = with > when the boundary line is dashed.

• replace = with ≥ when the boundary line is solid.

If the shaded half-plane is below the boundary line, then

• replace = with < when the boundary line is dashed.

• replace = with ≤ when the boundary line is solid.

Writing a Linear Inequality Using a Graph

Write an inequality that represents the graph.

SOLUTION

The boundar

5Solving Systems of Linear Equations - Big Ideas …...5Solving Systems of Linear Equations 5.1 Solving Systems of Linear Equations by Graphing 5.2 Solving Systems of Linear Equations

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